If the zeroes of (x^2-Sx+P) are (2\sqrt{3}+1) and (2\sqrt{3}-1), what are (S) and (P)?

(S=4\sqrt{3}), (P=11)

For a monic quadratic \(x^2-Sx+P\), the sum of the zeroes is \(S\), and their product is \(P\). The two zeroes are \(2\sqrt3+1\) and \(2\sqrt3-1\). Adding them cancels the 1 and \(-1\): \((2\sqrt3+1)+(2\sqrt3-1)=4\sqrt3\). Multiplying them uses the difference-of-squares identity: \((2\sqrt3)^2-1^2=12-1=11\).

Hence \(S=4\sqrt3\) and \(P=11\), which is exactly option A. The signs in the polynomial are important: in \(x^2-Sx+P\), the coefficient notation already makes the sum equal to \(S\), rather than \(-S\). The supplied answer is therefore correct and follows directly from the standard relations between roots and coefficients.

This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.